The base of a solid $S$ is the region bounded by the curve $y=e^{-x}$, the $x$ -axis, the $y$ -axis, and the line $x=1$. $(0,1)$ $x=1$ ${y=e^{-x}}$ $y$ $x$ Cross-sections perpendicular to the $x$ -axis are semi-circles. Determine the exact volume of solid $S$.
Explanation: Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $y$. $y$ $(x,y)$ $(0,1)$ $x=1$ ${y=e^{-x}}$ $y$ $x$ Since each cross-section is perpendicular to the $x$ -axis, the independent variable is $x$. If $A$ denotes the area of each cross-section as a function of $x$, the volume $V$ of solid $S$ is $ V=\int_a^b A(x) \,dx$. To determine the area $A$ as a function of $x$, first express $A$ in terms of $y$. Since the semi-circular cross-section rests on the rectangle pictured above, the diameter of the semi-circle is $y$. The radius of the semi-circle is $y/2$. $\dfrac y2$ $y$ The area $A$ of the semi-circle is $A=\dfrac12\cdot\pi \left(\dfrac y2\right)^2=\dfrac\pi8y^2$. What is $A$ as a function of $x$ ? The corner point $(x,y)$ of the rectangle lies on the curve $y=e^{-x}$. We can use this equation to express $A=\dfrac\pi8y^2$ in terms of $x$ as $A(x)=\dfrac\pi8(e^{-x})^2=\dfrac\pi8e^{-2x}$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $x$ goes from $0$ to $1$, the volume formula $ V=\int_a^b A(x) \,dx$ gives us the definite integral $\begin{aligned} V&=\int_0^1 \dfrac\pi8e^{-2x}\,dx \\\\ &=\dfrac\pi8\int_0^1 e^{-2x}\,dx \end{aligned}$ What is the value of the integral? $\begin{aligned} V&=\dfrac\pi8\int_0^1 e^{-2x}\,dx \\\\ &=\dfrac\pi8\left[-\dfrac12e^{-2x}\right]_0^1 \\\\ &=\dfrac\pi8\left[\dfrac12e^{-2x}\right]^0_1 \\\\ &=\dfrac\pi8\cdot\dfrac12\left[e^{-2(0)}-e^{-2(1)}\right] \\\\ &=\dfrac\pi{16}\left(1-e^{-2}\right) \end{aligned}$ Are you wondering what happened to the ${\text{minus sign}}$ in the coefficient of the antiderivative? We got rid of it by switching the order of the limits $0$ and $1$ ! This simple shortcut saves time and removes an opportunity to make a sign error.